crazy21482
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A hockey stick strikes a hockey puck of mass 0.17 kg. If the force exterted on the hockey puck is 35.0 N and there is a force of friction of 2.7 N opposing the motion. what is the acceleration of the hockey puck?

Choices:
0.0054 m/s^2
0.005 m/s^2
200 m/s^2
190 m/s^2

Respuesta :

Answer:

[tex]a=190\ m/s^2[/tex]

Explanation:

Mass of a hockey puck, m = 0.17 kg

Force exerted by the hockey puck, F' = 35 N

The force of friction, f = 2.7 N

We need to find the acceleration of the hockey puck.

Net force, F=F'-f

F=35-2.7

F=32.3 N

Now, using second law of motion,

F = ma

a is the acceleration of the hockey puck

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{32.3}{0.17}\\\\a=190\ m/s^2[/tex]

So, the acceleration of the hockey puck is [tex]190\ m/s^2[/tex].

snehashish65

The magnitude of acceleration of the hockey puck is 190 m/s² .

Given data:

The mass of hockey puck is, m = 0.17 kg.

the magnitude of external force on the puck is, F = 35.0 N.

The magnitude of force of friction is, f = 2.7 N.

We need to find the net force acting on the puck first. For that consider the applied force in the x-direction. Then net force is,

F' = F - f

Solving as,

F' = 35.0 - 2.7

F' = 32.3 N

Now, use the Newton's Second law to obtain the magnitude of acceleration (a) as,

F' = ma

Solving as,

32.3 = 0.17 (a)

a = 32.3/0.17

a = 190 m/s²

Thus, we can conclude that the magnitude of acceleration of the hockey puck is 190 m/s² .

Learn more about the Newton's law of motion here:

https://brainly.com/question/13447525

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